In fact, Guo and Schlosser [1]} have presented a \(q\) -supercongruence: for even \(d\ge 4\)  and integer \(n\ge 1\)  with \(n\equiv -1 \pmod {d}\) ,
\(\sum _{k=0}^{n-1}[2dk+1]\frac{(q;q^d)_{k}^d}{(q^d;q^d)_{k}^d}q^{\frac{d(d-3)k}{2}}\equiv 0 \pmod {\Phi _{n}(q)^2},\) 
